Assume a random variable $Z$ which is the sum of many non-observable random variables, say $Z = X_1 + \dotsm +X_n$ for which I know the probability density $p(X)$. Now given an observation $Z=z$, my problem is to find the most probable vector $(x_1,\dotsc,x_n)$ that generates that $z$. I would compute the mode of conditional joint probability density $p(x_1,x_2,\dotsc,x_n \mid Z=z)$. At the moment I need an expression for this conditional joint expression, all the material I have consulted so far didn't help much.

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    $\begingroup$ Using bayes rule, you can show that it is equal to 0 when $x_1 + x_2 + \ldots + x_n \neq z$ and equal to $\frac{\prod_i p(x_1)}{(\ast^n p)(z)}$ when the equality holds, where $(\ast^n p)$ is the shorthand for n-fold convolution of the marginal of $X$. $\endgroup$
    – Innuo
    Oct 15, 2015 at 14:04
  • $\begingroup$ @Innuo, are you assuming X are independent? How would change the formula if they are not? $\endgroup$
    – Davide C
    Oct 19, 2015 at 12:39

1 Answer 1


Since $z=x_1+\ldots+x_n$, the density of $(X_1,\ldots,X_{n-1},Z)$ is $$p(x_1)\cdots p(x_{n-1})p(z-x_1-\ldots-x_{n-1})$$ which is proportional to the conditional density of $(X_1,\ldots,X_{n-1}$ given $Z=z$, which is also the conditional density of $(X_1,\ldots,X_{n}$ given $Z=z$ since $X_n$ is a deterministic function of the others.


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